Question

What is the pH of 0.0050 HF (Ka = 6.8 x 10-4)?

2.73
11.70
11.27
2.30

Answer 1

The correct answer is 2.73.

HF is a weak acid which partially dissociates to release H+ and F-

HF → H⁺ + F⁻

Initial 0.0050 0 0

Change -x +x +x

Equilibrium 0.0050–x +x +x

Solve by using the equilibrium expression: = [H⁺] [F⁻]/ [HF]

6 .8 x 10⁻⁴= x. x / 0.0050 –x

6 .8 x 10⁻⁴= x² /0.0050

x² = 6 .8 x 10⁻⁴ x 0.0050

x² = 3.4 x 10⁻⁶

x = 3.4 x 10⁻⁶

[H⁺] = 1.84 x 10⁻³

pH = - log [H⁺] = - log (1.84 x 10⁻³)

pH = 2.73